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I What is the outer product of a tensor product of vectors?

  1. Apr 26, 2016 #1
    If one has two single-particle Hilbert spaces ##\mathcal{H}_{1}## and ##\mathcal{H}_{2}##, such that their tensor product ##\mathcal{H}_{1}\otimes\mathcal{H}_{2}## yields a two-particle Hilbert space in which the state vectors are defined as $$\lvert\psi ,\phi\rangle =\lvert\psi\rangle\otimes\lvert\phi\rangle\in\mathcal{H}_{1}\otimes\mathcal{H}_{2}$$ where ##\lvert\psi\rangle\in\mathcal{H}_{1}## and ##\lvert\phi\rangle\in\mathcal{H}_{2}##.

    Now, the inner product for ##\mathcal{H}_{1}\otimes\mathcal{H}_{2}## is defined such that $$\langle\phi ,\psi\vert\psi ,\phi\rangle =\left(\langle\phi\rvert\otimes\langle\psi\lvert\right)\left(\lvert\psi\rangle\otimes\lvert\phi\rangle\right) =\langle\psi\lvert\psi\rangle_{1}\langle\phi\lvert\phi\rangle_{2}$$ where ##\langle\cdot\lvert\cdot\rangle_{1}## is the inner product defined on ##\mathcal{H}_{1}## and ##\langle\cdot\lvert\cdot\rangle_{2}## the inner product defined on ##\mathcal{H}_{2}##.

    How though is the outer product defined? Is it simply $$\lvert\psi ,\phi\rangle\langle\phi ,\psi\rvert =\lvert\psi\rangle\otimes\lvert\phi\rangle\langle\phi\rvert\otimes\langle\psi\rvert =\lvert\psi\rangle\langle\psi\rvert_{1} \lvert\phi\rangle\langle\phi\rvert_{2}$$ where ##\lvert\psi\rangle\langle\psi\rvert_{1}## is the outer product in ##\mathcal{H}_{1}## and ##\lvert\phi\rangle\langle\phi\rvert_{2}## is the outer product in ##\mathcal{H}_{2}##.
     
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  3. Apr 26, 2016 #2

    andrewkirk

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    It might be easier to think of more general outer products, rather than the outer product of an element with itself, as you have written.
    That is, consider the outer product ##|a,b\rangle\langle c,d|=|a\rangle\otimes |b\rangle\langle c|\otimes\langle d|##.

    This is an operator on the product space ##\mathcal{H}_{1}\otimes\mathcal{H}_{2}##. To see what it does, we apply it to an element of that space ##|e,f\rangle=|e\rangle\otimes |f\rangle##.

    The result is ##|a,b\rangle \langle c|e\rangle \langle d|f\rangle##
     
  4. Apr 26, 2016 #3
    So one can't define it as ##\lvert a,b\rangle\langle c,d\rvert =\lvert a\rangle\langle c\rvert\lvert b\rangle\langle d\rvert## then? Should it be something like $$\lvert a,b\rangle\langle c,d\rvert =\lvert a\rangle\langle c\rvert\otimes\lvert b\rangle\langle d\rvert$$

    I ask in particular as I'm trying to understand the notion of a partial trace.
     
    Last edited: Apr 26, 2016
  5. Apr 26, 2016 #4

    andrewkirk

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    That symbol string on the RHS is not intrinsically meaningful. It juxtaposes the two operators ##\lvert a\rangle\langle c\rvert## and ##\lvert b\rangle\langle d\rvert## but:
    - since they are not scalars, the juxtaposition cannot be interpreted as scalar multiplication
    - since they are operators on different spaces (##\mathcal H_1## and ##\mathcal H_2##), the juxtaposition cannot be interpreted as composition of operation.

    That is one of a number of intrinsically meaningful ways of writing it. It has the same meaning as what I wrote above ##(
    |a\rangle\otimes |b\rangle)(\langle c|\otimes\langle d|)## (I have added parentheses here that were only implied above, to make it clear what operations are being performed).
     
  6. Apr 27, 2016 #5
    How does one show that the two expressions are equivalent? Would it be something like this:
    $$\left(\lvert a\rangle\otimes\lvert b\rangle\langle c\rvert\otimes\langle d\rvert\right)\lvert e\rangle\otimes\lvert f\rangle =\lvert a\rangle\otimes\lvert b\rangle\left(\langle c\vert e\rangle\langle d\vert f\rangle\right)\\ =\left(\lvert a\rangle\langle c\vert e\rangle\right)\otimes \left(\lvert b\rangle\langle d\vert f\rangle\right) =\left(\lvert a\rangle\langle c\lvert\otimes\lvert b\rangle\langle d\lvert\right)\lvert e\rangle\otimes\lvert f\rangle$$
     
  7. Apr 27, 2016 #6

    andrewkirk

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    I would write it as follows:

    $$(|a,b\rangle\langle c,d|)|e,f\rangle\equiv (|a\rangle\otimes|b\rangle)(\langle c|\otimes\langle d|)(|e\rangle\otimes|f\rangle)=(\langle c|e\rangle|a\rangle)\otimes (\langle d|f\rangle|b\rangle)$$
    and
    $$(|a\rangle\langle c|\otimes|b\rangle\langle d|)|e,f\rangle\equiv
    (|a\rangle\langle c|\otimes|b\rangle\langle d|)(|e\rangle\otimes |f\rangle)\
    =(|a\rangle\langle c|)|e\rangle\otimes (|b\rangle\langle d|)|f\rangle
    \equiv (\langle c|e\rangle|a\rangle)\otimes (\langle d|f\rangle|b\rangle)
    $$
    So
    $$(|a,b\rangle\langle c,d|)=(|a\rangle\langle c|\otimes|b\rangle\langle d|)$$
     
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